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Chapter 6: Problem 83

Calculate the work done (in joules) when 1.0 mole of water is frozen at \(0^{\circ} \mathrm{C}\) and \(1.0 \mathrm{~atm} .\) The volumes of one mole of water and ice at \(0^{\circ} \mathrm{C}\) are \(0.0180 \mathrm{~L}\) and \(0.0196 \mathrm{~L},\) respectively.

Short Answer

Expert verified
The work done on the system when 1 mole of water is frozen at \(0^{\circ} \mathrm{C}\) and \(1.0 \mathrm{~atm}\) is \(-0.162 \ \mathrm{J}\).

Step by step solution

01

Identify given quantities

In this exercise, several quantities are provided. We know the initial and final volumes of the water and ice respectively: \(V_{H_2O} = 0.0180 \ \mathrm{L}\) and \(V_{ice} = 0.0196 \ \mathrm{L}\). The pressure is also given, \(P = 1.0 \ \mathrm{atm}\). With these, we can calculate the work done during the freezing process.
02

Convert units

To apply the formula for work, \(w = \text{-P} \Delta V\), we need all quantities in respective SI units. The pressure should convert from atmospheres to pascals, and the volume from liters to cubic meters. By conversion, \(1 \ \text{atm} = 1.01325 \times 10^{5} \ \mathrm{Pa}\) and \(1 \ \mathrm{L} = 1.0 \times 10^{-3} \ \mathrm{m}^3\). Thus, \(P = 1.0 \ \mathrm{atm} = 1.01325 \times 10^{5} \ \mathrm{Pa}\), \(V_{H_2O} = 0.0180 \ \mathrm{L} = 0.0180 \times 10^{-3} \ \mathrm{m}^3\) and \(V_{ice} = 0.0196 \ \mathrm{L} = 0.0196 \times 10^{-3} \ \mathrm{m}^3\).
03

Calculate Change in Volume

The change in volume during the phase transition can be computed using the formula: \(\Delta V = V_{final} - V_{initial}\). Thus, \(\Delta V = V_{ice} - V_{H_2O} = 0.0196 \times 10^{-3} - 0.0180 \times 10^{-3} = 1.6 \times 10^{-6} \ \mathrm{m}^3\).
04

Use the Work Formula

The work done when water freezes at constant pressure is given by \(w=\text{-P} \Delta V\). Substituting the obtained values, \(w= -(1.01325 \times 10^{5} \ \mathrm{Pa})(1.6 \times 10^{-6} \ \mathrm{m}^3) = -0.162 \ \mathrm{J}\)
05

Interpret the result

The negative sign indicates that work is done by the system (the water) on the surroundings during the phase transition. The energy leaves the system, hence the work done is negative.

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Most popular questions from this chapter

The standard enthalpies of formation of ions in aqueous solutions are obtained by arbitrarily assigning a value of zero to \(\mathrm{H}^{+}\) ions; that is, \(\Delta H_{\mathrm{f}}^{\circ}\left[\mathrm{H}^{+}(a q)\right]=0\). (a) For the following reaction $$ \begin{aligned} \mathrm{HCl}(g) \stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{H}^{+}(a q) &+\mathrm{Cl}^{-}(a q) \\ \Delta H^{\circ} &=-74.9 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ calculate \(\Delta H_{\mathrm{f}}^{\circ}\) for the \(\mathrm{Cl}^{-}\) ions. (b) Given that \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{OH}^{-}\) ions is \(-229.6 \mathrm{~kJ} / \mathrm{mol}\), calculate the enthalpy of neutralization when 1 mole of a strong monoprotic acid (such as \(\mathrm{HCl}\) ) is titrated by 1 mole of a strong base (such as \(\mathrm{KOH}\) ) at \(25^{\circ} \mathrm{C}\)

A quantity of \(2.00 \times 10^{2} \mathrm{~mL}\) of \(0.862 \mathrm{M} \mathrm{HCl}\) is mixed with \(2.00 \times 10^{2} \mathrm{~mL}\) of \(0.431 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the \(\mathrm{HCl}\) and \(\mathrm{Ba}(\mathrm{OH})_{2}\) solutions is the same at \(20.48^{\circ} \mathrm{C}\). For the process $$ \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) $$ the heat of neutralization is \(-56.2 \mathrm{~kJ} / \mathrm{mol}\). What is the final temperature of the mixed solution?

Determine the amount of heat (in kJ) given off when \(1.26 \times 10^{4} \mathrm{~g}\) of \(\mathrm{NO}_{2}\) are produced according to the equation $$ \begin{aligned} 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow & 2 \mathrm{NO}_{2}(g) \\ \Delta H &=-114.6 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$

Explain what is meant by a state function. Give two examples of quantities that are state functions and two that are not.

Consider the following data: $$ \begin{array}{lcc} \text { Metal } & \text { Al } & \text { Cu } \\ \hline \text { Mass (g) } & 10 & 30 \\ \text { Specific heat }\left(\mathrm{J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right) & 0.900 & 0.385 \\ \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & 40 & 60 \end{array} $$ When these two metals are placed in contact, which of the following will take place? (a) Heat will flow from Al to Cu because Al has a larger specific heat. (b) Heat will flow from Cu to Al because Cu has a larger mass. (c) Heat will flow from Cu to Al because Cu has a larger heat capacity. (d) Heat will flow from Cu to Al because Cu is at a higher temperature. (e) No heat will flow in either direction.
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